Радиус окружности в треугольнике

Analyzing the quadrilateral

I’m thinking through quadrilateral $PQRS$ with intersecting diagonals at $O$. Given conditions are: $PR \perp QS$, $PS \parallel QR$, $QR = 91$, $OS = 12$, and the inradius of $\triangle ORS$ is 5. The task is to find the inradius of $\triangle OPS$.

Piecing together

I’m working through the given elements: $PR \perp QS$, $PS \parallel QR$, $QR = 91$, $OS = 12$, and the inradius of $\triangle ORS$ is 5. The goal is to find the inradius of $\triangle OPS$.

Considering geometric implications

I’m gathering info on the shape possibilities of quadrilateral $PQRS$. With $PS \parallel QR$, it might be a trapezoid or isosceles trapezoid, leading to interesting geometric insights.

Applying geometric principles

I’m using the formula for the incircle radius of a triangle to relate areas and perimeters of $\triangle ORS$ and $\triangle OPS$, considering their geometric properties and reciprocal relations.

Placing the quadrilateral

Hmm, I’m thinking about mapping quadrilateral $PQRS$ on a coordinate plane, setting $O$ at the origin, $R$ on the x-axis, $P$ on the negative x-axis, $Q$ on the positive y-axis, and $S$ on the negative y-axis.

Setting coordinates

Starting with $O$ at $(0,0)$ and $R$ on the positive x-axis as $(r, 0)$. Considering $Q$ at $(0, q)$ and $S$ at $(0, -12)$ to satisfy $OS = 12$.

Analyzing coordinates

I’m positioning $R$ on the x-axis and $P$ on the same line. Using $QR$ length, I’m ensuring $PS$ is parallel to $QR$.

Ensuring parallelism

I’m working through the parallelism of $PS$ and $QR$ by setting the ratio of their vectors equal. This leads to equations $\frac{x_R}{-x_P} = \frac{q}{12}$ to verify.

Confirming parallelism

I’m verifying $PS$ is parallel to $QR$ by ensuring the vector ratio $\frac{x_R}{x_P} = -\frac{q}{12}$ is accurate, consistent with the known lengths.

Calculating triangle dimensions

OK, let's determine the inradius of $\triangle ORS$ with vertices $(0,0)$, $(x_R,0)$, and $(0,-12)$. The sides are $OR = x_R$, $OS = 12$, and $RS = \sqrt{x_R^2 + 144}$.

Calculating inradius

Taking a closer look at $\triangle ORS$, which has an area of $6x_R$, a perimeter of $x_R + 12 + \sqrt{x_R^2 + 144}$, and a semiperimeter to find the inradius.

Solving for $x_R$

Using the inradius formula for $\triangle ORS$, I substituted the known values and tackled the equation to find $x_R$, positioning $R$ at $(35,0)$.

Finding $q$

I’m solving for $q$ by noting the length $QR = 91$ and applying the distance formula to determine $q = \sqrt{7056}$, confirming values for $R$ and $S$.

Determining vector ratios

I’m working through parallel vectors to find $P$'s coordinates by ensuring the ratio $\frac{35}{-x_P} = \frac{-84}{-12}$, aiming to locate $P$ and ultimately the inradius of $\triangle OPS$.

Pinpointing P's coordinates

I set up the ratio $\frac{35}{-x_P} = \frac{-84}{-12}$, which led to $k = 7$. Calculating $x_P$ as $-5$, I confirmed $P(-5,0)$.

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